How to calculate efficiency of Newton's Method

[David Carroll]

Greetings. I was wondering if anyone knew of a way to calculate the efficiency of Newton's Method for a given function:

I have an equation f(x) and I'm trying to find a value of x = x₀ such that f(x₀) = 0.

So I start with a guess x₀ and then use that to find a second (usually closer) guess of x₁ by using the following:

x₁ = x₀ - f(x₀)/f'(x₀)

and then a third guess

x₂ = x₁ - f(x₁)/f'(x₁)

...and so on until x_n = x_n-1 +- e, where "e" is some previously defined allowable error.

In order to find an "efficiency value" of the above method, I'm thinking this has to do with the rate at which f(x) changes versus the rate at which f"(x) changes. So would I take the derivative of f(x) and divide it by the derivative of f'(x)? = f'(x)/f''(x)

...or would I take the derivative of the entire construct? = [f(x)/f'(x)]'

Or are none of these correct?

 

[David Carroll]

Sorry. You can barely see the ' or the '' on those above derivatives, but they are there.

 

[SteamKing]

Greetings. I was wondering if anyone knew of a way to calculate the efficiency of Newton's Method for a given function:

I have an equation f(x) and I'm trying to find a value of x = x₀ such that f(x₀) = 0.

So I start with a guess x₀ and then use that to find a second (usually closer) guess of x₁ by using the following:

x₁ = x₀ - f(x₀)/f'(x₀)

and then a third guess

x₂ = x₁ - f(x₁)/f'(x₁)

...and so on until x_n = x_n-1 +- e, where "e" is some previously defined allowable error.

In order to find an "efficiency value" of the above method, I'm thinking this has to do with the rate at which f(x) changes versus the rate at which f"(x) changes. So would I take the derivative of f(x) and divide it by the derivative of f'(x)? = f'(x)/f''(x)

...or would I take the derivative of the entire construct? = [f(x)/f'(x)]'

Or are none of these correct?

It's not clear what you mean by the 'efficiency' of Newton's method.

http://en.wikipedia.org/wiki/Newton's_method

In the article above, it can be shown that under certain conditions, NM has what is called 'quadratic convergence', which means that if NM has settled onto a solution, the error in a subsequent iteration is proportional to the square of the error in the current iteration.

 

[David Carroll]

That's exactly what I was looking for. Thank you.